The figure above shows the region A, which is bounded by the x- and y-axes, the graph of $f(x)=\frac{\sin x}{x}$

Key concept: This is a related-rates problem for an area defined by a moving boundary. The area depends on $k$, so the rate of change of the area comes from differentiating the integral with respect to the moving endpoint and then multiplying by $dk/dt$.

Step by step

Let the area be

$A(k)=\int_0^k \frac{\sin x}{x}\,dx.$

By the Fundamental Theorem of Calculus,

$\frac{dA}{dk}=\frac{\sin k}{k}.$

Since $k$ changes with time at rate

$\frac{dk}{dt}=\frac{\pi}{4},$

use the chain rule:

\frac{\sin k}{k}\cdot \frac{\pi}{4}.$$ At $k=\pi/6$, $$\frac{\sin(\pi/6)}{\pi/6}=\frac{1/2}{\pi/6}=\frac{3}{\pi}.$$ So $$\frac{dA}{dt}=\frac{3}{\pi}\cdot \frac{\pi}{4}=\frac{3}{4}.$$ Thus the area is increasing at $$\boxed{\frac{3}{4}}$$ so the correct choice is **B**. ### Pitfall alert Do not plug $k=\pi/6$ directly into the area formula as if the answer were just the area itself. The question asks for a rate, so you need both the derivative with respect to $k$ and the rate $dk/dt$. Also, make sure to use $\sin k/k$ at the endpoint, not an average value over the whole interval. ### Try different conditions If $k$ were constant, then $dk/dt=0$ and the area would not change with time. If the boundary function were $\frac{\sin(2x)}{x}$ instead, the same method would still apply, but the endpoint evaluation would change accordingly. ### Further reading related rates, Fundamental Theorem of Calculus, chain rule